M18 Exponential and Logarithmic Functions.pdf

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Exponential and

Logarithmic Functions

At the end of this lecture, a student must be able to:

Demonstrate the properties of real exponents

Recognize an exponential function and its properties

Solve equations involving expressions with real exponents Illustrate the relation between a logarithm and an

expression involving exponents

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Some Applications of Exponential and

Logarithmic Functions

•

Exponential growth and decay

•

Compound interest

•

Acidity of chemical substances

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Real Exponents

recall: definition and properties of rational exponents

Ifa >1 and p, q ∈_Q with p < q, then ap < aq.

Example: 23 _<₂4_{, 3}−2 _<₃−1_{, 5}1.3 _<₅1.32

We want to define 2

√

(4)

Real Exponents

Ifa >1 and p, q ∈_Q with p < q,

then ap < aq.

Example: 23 _<₂4_{, 3}−2 _<₃−1_{, 5}1.3 _<₅1.32

We want to define 2

√

(5)

Real Exponents

Ifa >1 and p, q ∈_Q with p < q, then ap _{< a}q_.

Example: 23 _<₂4_{, 3}−2 _<₃−1_{, 5}1.3 _<₅1.32

We want to define 2

√

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Real Exponents

Example: 23 _<₂4_,

3−2 _<₃−1_{, 5}1.3 _<₅1.32

We want to define 2

√

(7)

Real Exponents

Example: 23 _<₂4_{, 3}−2 _<₃−1_,

51.3 _<₅1.32

We want to define 2

√

(8)

Real Exponents

Example: 23 _<₂4_{, 3}−2 _<₃−1_{, 5}1.3 _<₅1.32

We want to define 2

√

(9)

Real Exponents

Example: 23 _<₂4_{, 3}−2 _<₃−1_{, 5}1.3 _<₅1.32

We want to define 2

√

(10)

Real Exponents

Note: √2≈1.41421359....

If 2

√

2 _{is to be defined such that properties of exponents} would hold,

Then:

21 <2

√

2

21.4 <2

√

2

21.41<2

√

2

21.414<2

√

2

...

At the same time,

2

√

2 _<₂2

2

√

2 _<₂1.5

2

√

2 _<₂1.42

2

√

2 _<₂1.415

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Real Exponents

Note: √2≈1.41421359....

If 2

√

Then:

21 <2

√

2

21.4 <2

√

2

21.41<2

√

2

21.414<2

√

2

...

At the same time,

2

√

2 _<₂2

2

√

2 _<₂1.5

2

√

2 _<₂1.42

2

√

2 _<₂1.415

(12)

Real Exponents

Note: √2≈1.41421359....

If 2

√

Then:

21 <2

√

2

21.4 <2

√

2

21.41<2

√

2

21.414<2

√

2

...

At the same time,

2

√

2 _<₂2

2

√

2 _<₂1.5

2

√

2 _<₂1.42

2

√

2 _<₂1.415

(13)

Real Exponents

Note: √2≈1.41421359....

If 2

√

Then:

21 <2

√

2

21.4 <2

√

2

21.41<2

√

2

21.414<2

√

2

...

At the same time,

2

√

2 _<₂2

2

√

2 _<₂1.5

2

√

2 _<₂1.42

2

√

2 _<₂1.415

(14)

Real Exponents

Note: √2≈1.41421359....

If 2

√

Then:

21 <2

√

2

21.4 <2

√

2

21.41<2

√

2

21.414 <2

√

2

...

At the same time,

2

√

2 _<₂2

2

√

2 _<₂1.5

2

√

2 _<₂1.42

2

√

2 _<₂1.415

(15)

Real Exponents

Note: √2≈1.41421359....

If 2

√

Then:

21 <2

√

2

21.4 <2

√

2

21.41<2

√

2

21.414 <2

√

2

...

At the same time,

2

√

2 _<₂2

2

√

2 _<₂1.5

2

√

2 _<₂1.42

2

√

2 _<₂1.415

(16)

Real Exponents

Note: √2≈1.41421359....

If 2

√

Then:

21 <2

√

2

21.4 <2

√

2

21.41<2

√

2

21.414 <2

√

2

...

At the same time,

2

√

2 _<₂2

2

√

2 _<₂1.5

2

√

2 _<₂1.42

2

√

2 _<₂1.415

(17)

Real Exponents

Note: √2≈1.41421359....

If 2

√

Then:

21 <2

√

2

21.4 <2

√

2

21.41<2

√

2

21.414 <2

√

2

...

At the same time,

2

√

2 _<₂2

2

√

2 _<₂1.5

2

√

2 _<₂1.42

2

√

2 _<₂1.415

(18)

Real Exponents

Note: √2≈1.41421359....

If 2

√

Then:

21 <2

√

2

21.4 <2

√

2

21.41<2

√

2

21.414 <2

√

2

...

At the same time,

2

√

2 _<₂2

2

√

2 _<₂1.5

2

√

2 _<₂1.42

2

√

2 _<₂1.415

(19)

Real Exponents

Strictly speaking (though we will not dwell on this),

Definition

Ifa >1 and r∈_R, ar _{is defined as the}_{least upper bound} _of

{aq |q ≤r, q∈_Q}.

Definition

If 0< a <1 and r∈_R, ar _{is defined as the} _{greatest lower}

bound of

(20)

Properties of Real Exponents

Theorem

Let a, b, x, y∈_R and a, b > 0,

1. ax is a unique real number.

2. a0 = 1.

3. if a= 1, then ax _{= 1}_.

4. a−x ₌ 1 ax

5. laws of real exponents

a. axay =ax+y

b. a x

ay =a x−y

c. (ab)x =axbx

d. a

b

x

= a

x

bx

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Theorem

6. if a >1 with x < y then ax _{< a}y

7. if 0< a < 1 with x < y then ax _{> a}y

Example:

(6) 2<3 and 42 _<₄3

(7) 2<3 but 1 2

2

>1

2

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Properties of Real Exponents

Theorem

Example:

(6) 2<3 and 42 _<₄3

(7) 2<3 but 1 2

2

>1

2

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Theorem

Example:

(6) 2<3 and 42 _<₄3

(7) 2<3 but 1 2

2

>1

2

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Theorem

Example:

(6) 2<3 and 42 _<₄3

(7) 2<3 but 1 2

2

>1

2

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Theorem

Example:

(6) 2<3 and 42 _<₄3

(7) 2<3 but 1 2

2

>1

2

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Exponential Functions

Definition

Ifb >0, b6= 1, theexponential function with base b is defined by

f(x) =bx

for every x∈_R.

Examples:

(1) f(x) = 2x

(2) f(x) = 1₂x

(3) f(x) =πx

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Exponential Functions

Definition

Ifb >0, b6= 1, theexponential function with base b is defined by

f(x) =bx

for every x∈_R.

Examples:

(1) f(x) = 2x

(2) f(x) = 1₂x

(3) f(x) =πx

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Graphs of Exponential Functions

Note: Forb >0 and b 6= 1, bx is always a positive real number.

Letb >0, b6= 1 andf be the exponential function with base b.

1. domf =_R

2. ranf = (0,+∞)

3. x-int: none

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Graphs of Exponential Functions

1. domf =_R

2. ranf = (0,+∞)

3. x-int: none

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Graphs of Exponential Functions

1. domf =

R

2. ranf = (0,+∞)

3. x-int: none

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1. domf =_R

2. ranf = (0,+∞)

3. x-int: none

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Graphs of Exponential Functions

1. domf =_R

2. ranf =

(0,+∞)

3. x-int: none

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1. domf =_R

2. ranf = (0,+∞)

3. x-int: none

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Graphs of Exponential Functions

1. domf =_R

2. ranf = (0,+∞)

3. x-int:

none

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1. domf =_R

2. ranf = (0,+∞)

3. x-int: none

(36)

Graphs of Exponential Functions

1. domf =_R

2. ranf = (0,+∞)

3. x-int: none

4. y-int:

(37)

1. domf =_R

2. ranf = (0,+∞)

3. x-int: none

(38)

Letb >1 and f be the exponential function with base b.

5. Ifx1 < x2,

then bx1 _{< b}x2_.

(f is an increasing function)

6. f(x) = bx _{is one-to-one}

−4. −3. −2. −1. 1. 2. 3. 1.

2. 3. 4.

0

y= 2x

(39)

5. Ifx1 < x2, then bx1 < bx2.

(f is an increasing function)

−4. −3. −2. −1. 1. 2. 3. 1.

2. 3. 4.

0

y= 2x

(40)

5. Ifx1 < x2, then bx1 < bx2. (f is an increasing function)

−4. −3. −2. −1. 1. 2. 3. 1.

2. 3. 4.

0

y= 2x

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6. f(x) =bx _{is one-to-one}

−4. −3. −2. −1. 1. 2. 3. 1.

2. 3. 4.

0

y= 2x

(42)

−4. −3. −2. −1. 1. 2. 3. 1.

2. 3. 4.

0

y= 2x

(43)

Let0< b <1 and f be the exponential function with base

b.

5. Ifx1 < x2,

then bx1 _{> b}x2_.

(f is a decreasing function)

−3. −2. −1. 1. 2. 3. 4. 1.

2. 3. 4.

(44)

b.

5. Ifx1 < x2, then bx1 > bx2.

(f is a decreasing function)

−3. −2. −1. 1. 2. 3. 4. 1.

2. 3. 4.

(45)

b.

5. Ifx1 < x2, then bx1 > bx2. (f is a decreasing function)

−3. −2. −1. 1. 2. 3. 4. 1.

2. 3. 4.

(46)

b.

−3. −2. −1. 1. 2. 3. 4. 1.

2. 3. 4.

(47)

b.

−3. −2. −1. 1. 2. 3. 4. 1.

2. 3. 4.

(48)

The Number

e

e= 2.718281828459045...

Definition

The natural exponential functionis the exponential function with basee: f(x) =ex

−4. −3. −2. −1. 1. 2. 3. 1.

2. 3.

0 (0,1)

(49)

The Number

e

e= 2.718281828459045...

Definition

The natural exponential functionis the exponential function with basee: f(x) =ex

−4. −3. −2. −1. 1. 2. 3. 1.

2. 3.

0 (0,1)

(50)

Equations involving Exponential Expressions

Letb >0, b6= 1.

f(x) = bx _{is one-to-one} m

if br =bs then r=s.

Example: 53x _{= 5}7x−2 Solution:

3x= 7x−2

2 = 4x

x= 1

(51)

Letb >0, b6= 1.

if br =bs then r=s.

Example: 53x _{= 5}7x−2

Solution:

3x= 7x−2

2 = 4x

x= 1

(52)

Equations involving Exponential Expressions

Letb >0, b6= 1.

if br =bs then r=s.

3x= 7x−2

2 = 4x

x= 1

(53)

Letb >0, b6= 1.

if br =bs then r=s.

3x= 7x−2

2 = 4x

x= 1

(54)

Letb >0, b6= 1.

if br =bs then r=s.

3x= 7x−2

2 = 4x

x= 1

(55)

Example: 4t2

= 46−t

Solution:

t2 = 6−t t2 +t−6 = 0 (t+ 3)(t−2) = 0

(56)

Example: 4t2

= 46−t Solution:

t2 = 6−t

t2 +t−6 = 0 (t+ 3)(t−2) = 0

(57)

Example: 4t2

= 46−t Solution:

t2 = 6−t t2+t−6

= 0 (t+ 3)(t−2) = 0

(58)

Example: 4t2

= 46−t Solution:

t2 = 6−t t2+t−6 = 0

(t+ 3)(t−2) = 0

(59)

Example: 4t2

= 46−t Solution:

t2 = 6−t t2+t−6 = 0 (t+ 3)

(t−2) = 0

(60)

Example: 4t2

= 46−t Solution:

t2 = 6−t t2+t−6 = 0 (t+ 3)(t−2)

= 0

(61)

Example: 4t2

= 46−t Solution:

t2 = 6−t t2+t−6 = 0 (t+ 3)(t−2) = 0

(62)

Example: 4t2

= 46−t Solution:

t2 = 6−t t2+t−6 = 0 (t+ 3)(t−2) = 0

t =−3 or

(63)

Example: 4t2

= 46−t Solution:

t2 = 6−t t2+t−6 = 0 (t+ 3)(t−2) = 0

(64)

Example: 3z = 9z+5

Solution:

Express in terms of the same base:

3z = (32)z+5

3z = 32z+10

z = 2z+ 10

(65)

Example: 3z = 9z+5 Solution:

3z = (32)z+5

3z = 32z+10

z = 2z+ 10

(66)

3z

= (32)z+5

3z = 32z+10

z = 2z+ 10

(67)

3z = (32)z+5

3z = 32z+10

z = 2z+ 10

(68)

3z = (32)z+5

3z

= 32z+10

z = 2z+ 10

(69)

3z = (32)z+5

3z = 32z+10

z = 2z+ 10

(70)

3z = (32)z+5

3z = 32z+10

z

= 2z+ 10

(71)

3z = (32)z+5

3z = 32z+10

z = 2z+ 10

(72)

3z = (32)z+5

3z = 32z+10

z = 2z+ 10

(73)

Example: 45−9x ₌ 1 8x−2

Solution:

22(5−9x)= 2−3(x−2) 2(5−9x) = −3(x−2)

10−18x=−3x+ 6 4 = 15x

x= 4

(74)

Example: 45−9x ₌ 1 8x−2 Solution:

2

2(5−9x)

= 2−3(x−2) 2(5−9x) = −3(x−2)

10−18x=−3x+ 6 4 = 15x

x= 4

(75)

22(5−9x)

= 2−3(x−2) 2(5−9x) = −3(x−2)

10−18x=−3x+ 6 4 = 15x

x= 4

(76)

22(5−9x)= 2

−3(x−2)

2(5−9x) = −3(x−2) 10−18x=−3x+ 6

4 = 15x

x= 4

(77)

22(5−9x)= 2−3(x−2)

2(5−9x) = −3(x−2) 10−18x=−3x+ 6

4 = 15x

x= 4

(78)

22(5−9x)= 2−3(x−2) 2(5−9x) = −3(x−2)

10−18x=−3x+ 6 4 = 15x

x= 4

(79)

22(5−9x)= 2−3(x−2) 2(5−9x) = −3(x−2)

10−18x=−3x+ 6

4 = 15x

x= 4

(80)

22(5−9x)= 2−3(x−2) 2(5−9x) = −3(x−2)

10−18x=−3x+ 6 4 = 15x

x= 4

(81)

22(5−9x)= 2−3(x−2) 2(5−9x) = −3(x−2)

10−18x=−3x+ 6 4 = 15x

x= 4

(82)

Example: 9x_{+ 2 (3}x₎₋_{3 = 0}

Solution:

32x+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2 + 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

3x =−3 has no solution since 3x _>₀

(83)

Example: 9x_{+ 2 (3}x₎₋_{3 = 0} Solution:

3

2x

+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2 + 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

(84)

32x

+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2 + 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

(85)

32x+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2 + 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

(86)

32x+ 2(3x)−3 = 0

(3x)2

+ 2(3x)−3 = 0 Let y= 3x

y2 + 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

(87)

32x+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0

Let y= 3x

y2 + 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

(88)

32x+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2 + 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

(89)

32x+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2+ 2y−3 = 0

(y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

(90)

32x+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2+ 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

(91)

32x+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2+ 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3

3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

(92)

32x+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2+ 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

(93)

32x+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2+ 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒

x= 0

(94)

32x+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2+ 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

(95)

32x+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2+ 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0 3x =−3

has no solution since 3x >0

(96)

32x+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2+ 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

(97)

32x+ 2(3x)−3 = 0

(3x)2+ 2(3x)−3 = 0 Let y= 3x

y2+ 2y−3 = 0 (y−1)(y+ 3) = 0

y= 1 or y=−3 3x= 1 or 3x =−3

3x = 1 ⇒ x= 0

(98)

Logarithms

Definition

Letb∈_R such that b >0 and b6= 1. If by ₌_x _then _y _is called the logarithm ofx to the base b, denoted y= log_bx.

Examples: Let a∈_R, a >0 and a6= 1.

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(99)

Logarithms

Definition

Letb∈_R such that b >0 and b6= 1. If by ₌_x _then _y _is called the logarithm ofx to the base b, denoted y= log_bx. Examples: Let a∈_R,a >0 and a6= 1.

(1) log₄16 =

2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(100)

Logarithms

Definition

(1) log₄16 =2

because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(101)

Logarithms

Definition

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(102)

Logarithms

Definition

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 =

−3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(103)

Logarithms

Definition

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3

because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(104)

Logarithms

Definition

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(105)

Logarithms

Definition

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 =

−4 because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(106)

Logarithms

Definition

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4

because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(107)

Logarithms

Definition

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(108)

Logarithms

Definition

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 =

0 since a0 _{= 1}

(109)

Logarithms

Definition

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 = 0

since a0 _{= 1}

(110)

Logarithms

Definition

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(111)

Logarithms

Definition

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(5) log_aa=

(112)

Logarithms

Definition

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(5) log_aa= 1

(113)

Logarithms

Definition

(1) log₄16 =2 because 42 _{= 16}

(2) log₅ ₁₂₅1 = −3 because 5−3 = ₁₂₅1

(3) log1 3

81 = −4 because 1₃−4 = 81

(4) log_a1 = 0 since a0 _{= 1}

(114)

Computing the Inverse of the Exponential

Function

Letb∈_R such that b >0 and b6= 1.

Forf(x) = bx, solve for f−1(x).

Interchangingx and y, we have

x=by m log_bx=y

Thus,

(115)

Computing the Inverse of the Exponential

Function

Forf(x) = bx, solve for f−1(x). Interchanging x and y, we have

x=by m log_bx=y

Thus,

(116)

Function

x=by

m log_bx=y

Thus,

(117)

Computing the Inverse of the Exponential

Function

x=by m log_bx=y

Thus,

(118)

Function

x=by m log_bx=y

Thus,

(119)

Definition

Letb∈_R such that b >0 and b6= 1. The function

f(x) = log_bx

is called thelogarithmic function to the base b.

Notes:

1 The logarithmic function to the base b and exponential function with baseb are inverse functions of each other.

2 The domain of f(x) = log_bx is (0,+∞). Its range is_R.

3 log_b(bx_{) =} _x _{for all} _x_∈

R

(120)

Definition

f(x) = log_bx

is called thelogarithmic function to the base b. Notes:

R

(121)

Definition

f(x) = log_bx

2 The domain of f(x) = log_bx is

(0,+∞). Its range is_R.

R

(122)

Definition

f(x) = log_bx

2 The domain of f(x) = log_bx is (0,+∞).

Its range is _R.

R

(123)

Definition

f(x) = log_bx

2 The domain of f(x) = log_bx is (0,+∞). Its range is

R.

R

(124)

Definition

f(x) = log_bx

R

(125)

Definition

f(x) = log_bx

3 log_b(bx_{) =}

x for all x∈_R

(126)

Definition

f(x) = log_bx

R

(127)

Definition

f(x) = log_bx

R

4 blogbx ₌

(128)

Definition

f(x) = log_bx

R

(129)

Graphs of Logarithmic Functions

(0,1)

(1,0)

b >1

(0,1)

(1,0) 0< b <1

(1,0)

b >1

0< b <1

Notes:

5 x-int: 1

(130)

Graphs of Logarithmic Functions

(0,1)

(1,0)

b >1

(0,1)

(1,0) 0< b <1

(1,0)

b >1

0< b <1

Notes:

5 x-int: 1

(131)

(0,1)

(1,0) b >1

(0,1)

(1,0) 0< b <1

(1,0)

b >1

0< b <1

Notes:

5 x-int: 1

(132)

(0,1)

(1,0) b >1

(0,1)

(1,0)

0< b <1

(1,0)

b >1

0< b <1

Notes:

5 x-int: 1

(133)

(0,1)

(1,0) b >1

(0,1)

(1,0)

0< b <1

(1,0)

b >1

0< b <1

Notes:

5 x-int: 1

(134)

(0,1)

(1,0) b >1

(0,1)

(1,0) 0< b <1

(1,0)

b >1

0< b <1

Notes:

5 x-int: 1

(135)

(0,1)

(1,0) b >1

(0,1)

(1,0) 0< b <1

(1,0)

b >1

0< b <1

Notes:

5 x-int: 1

(136)

Graphs of Logarithmic Functions

(0,1)

(1,0) b >1

(0,1)

(1,0) 0< b <1

(1,0)

b >1

0< b <1

Notes:

5 x-int: 1

(137)

(0,1)

(1,0) b >1

(0,1)

(1,0) 0< b <1

(1,0)

b >1

0< b <1

Notes:

5 x-int: 1

(138)

(0,1)

(1,0) b >1

(0,1)

(1,0) 0< b <1

(1,0)

b >1

0< b <1

Notes:

5 x-int: 1

(139)

(1,0)

b >1

0< b <1

Notes:

7 Ifb >1, it is an increasing function. Moreover, it is positive when x >1 and negative when 0< x <1.

(140)

(1,0)

b >1

0< b <1

Notes:

7 Ifb >1, it is an increasing function. Moreover, it is positive whenx >1 and negative when 0< x <1.

(141)

Graphs of Logarithmic Functions

(1,0)

b >1

0< b <1

Notes:

(142)

(1,0)

b >1

0< b <1

Notes:

(143)

Common and Natural Logarithms

Definition

Letx∈_R such thatx >0.

• The common logarithm of x, denoted logx, is

logx= log₁₀x.

• The natural logarithm of x, denoted lnx, is

(144)

Common and Natural Logarithms

Definition

Letx∈_R such thatx >0.

• The common logarithm of x, denoted logx, is

logx= log₁₀x.

• The natural logarithm of x, denoted lnx, is

(145)

Recall: log_bx=y⇔by ₌_x

Examples:

(1) log 100 =

2

(2) log 1

1000 = −3

(3) ln√3 _e₌ 1 3

(4) ln1

(146)

Examples:

(1) log 100 = 2

(2) log 1

1000 = −3

(3) ln√3 _e₌ 1 3

(4) ln1

(147)

Common and Natural Logarithms

Examples:

(1) log 100 = 2

(2) log 1 1000 =

−3

(3) ln√3 _e₌ 1 3

(4) ln1

(148)

Examples:

(1) log 100 = 2

(2) log 1

1000 = −3

(3) ln√3 _e₌ 1 3

(4) ln1

(149)

Examples:

(1) log 100 = 2

(2) log 1

1000 = −3

(3) ln√3 _e₌

1 3

(4) ln1

(150)

Examples:

(1) log 100 = 2

(2) log 1

1000 = −3

(3) ln√3 _e₌ 1 3

(4) ln1

(151)

Common and Natural Logarithms

Examples:

(1) log 100 = 2

(2) log 1

1000 = −3

(3) ln√3 _e₌ 1 3

(4) ln1

e =

(152)

Examples:

(1) log 100 = 2

(2) log 1

1000 = −3

(3) ln√3 _e₌ 1 3

(4) ln1

(153)

Common and Natural Logarithmic Functions

Definition

Letx∈_R with x >0.

1. The common logarithmic functionis defined by

f(x) = logx.

2. The natural logarithmic function is defined by

f(x) = lnx.

Note: The domain of the common and natural logarithmic

(154)

Common and Natural Logarithmic Functions

Definition

f(x) = logx.

f(x) = lnx.

Note: The domain of the common and natural logarithmic function is

(155)

Common and Natural Logarithmic Functions

Definition

f(x) = logx.

f(x) = lnx.

(156)

Common and Natural Logarithmic Functions

Definition

f(x) = logx.

f(x) = lnx.

(157)

Sincef−1_◦_f₍_x_{) =} _x _{for all} _x_∈_dom_f_{, we have..}

1. log 10x

=x for every x∈_R

2. 10logx =x for every x∈(0,+∞)

3. lnex ₌_x _{for every} _x_∈

R

(158)

1. log 10x =x for every x∈_R

2. 10logx =x for every x∈(0,+∞)

R

(159)

2. 10logx

=x for every x∈(0,+∞)

R

(160)

2. 10logx ₌_x _{for every} _x_∈₍₀_,_+∞)

R

(161)

2. 10logx ₌_x _{for every} _x_∈₍₀_,_+∞)

3. lnex

=x for every x∈_R

(162)

2. 10logx ₌_x _{for every} _x_∈₍₀_,_+∞)

R

(163)

2. 10logx ₌_x _{for every} _x_∈₍₀_,_+∞)

R

4. elnx

(164)

2. 10logx ₌_x _{for every} _x_∈₍₀_,_+∞)

R

(165)

Recap:

Demonstrate the properties of real exponents

Recognize an exponential function and its properties

Solve equations involving expressions with real exponents

Illustrate the relation between a logarithm and an expression involving exponents

(166)

Exercises:

1. Solve forx

1.1 23x = ₆₄1

1.2 31−x _{= 9}2x 1.3 25x+2 = ₁₂₅11−x 1.4 16x2−1−8x−1= 0

1.5 25x−6(5x) + 5 = 0

1.6 (3x)2−10(3x) + 9 = 0

1.7 22x+1+ 4x = 24

1.8 2x+ 2−x = 2

2. Express the following in logarithmic form

2.1 82 = 64 2.2 2−5 = ₃₂1 2.3 1634 = 8

3. Find the value of 28x if 16x = 5.

4. Find the respective domains and ranges of the functions defined byf(x) = log₅(x+ 2) + 3 and